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1372. Longest ZigZag Path in a Binary Tree

Description

You are given the root of a binary tree.

A ZigZag path for a binary tree is defined as follow:

  • Choose any node in the binary tree and a direction (right or left).
  • If the current direction is right, move to the right child of the current node; otherwise, move to the left child.
  • Change the direction from right to left or from left to right.
  • Repeat the second and third steps until you can't move in the tree.

Zigzag length is defined as the number of nodes visited - 1. (A single node has a length of 0).

Return the longest ZigZag path contained in that tree.

 

Example 1:

Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1]
Output: 3
Explanation: Longest ZigZag path in blue nodes (right -> left -> right).

Example 2:

Input: root = [1,1,1,null,1,null,null,1,1,null,1]
Output: 4
Explanation: Longest ZigZag path in blue nodes (left -> right -> left -> right).

Example 3:

Input: root = [1]
Output: 0

 

Constraints:

  • The number of nodes in the tree is in the range [1, 5 * 104].
  • 1 <= Node.val <= 100

Solutions

Solution 2: Explicit Stack

Thinking

A zigzag must alternate left and right as it descends. Carrying the length of a path that last stepped left ($l$) or right ($r$), then recursing, is enough on a short tree: the left child continues with $r+1$ and a zero right length, and the right child is symmetric.

The tree can contain $5\times 10^4$ nodes. A chain recurses once per node and overflows the call stack.

The lengths at a node depend only on the parent’s other direction, and the answer is a running maximum, so the walk does not need a return value.

An explicit stack stores each node with the $l$ and $r$ of the step that reached it. After a node is popped, the maximum is updated and its children are pushed with those new lengths. The left child is pushed last, so it is visited first.

We walk the tree with an explicit stack. Each frame stores the current node and the zigzag lengths $l$ and $r$ of the step that arrived there.

After a node is popped, $\max(l, r)$ updates the answer. A left child is pushed with left length $r+1$ and right length $0$; a right child is pushed with right length $l+1$ and left length $0$. The root starts with both lengths equal to $0$.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the number of nodes.

  • /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int ans;

    public int longestZigZag(TreeNode root) {
        dfs(root, 0, 0);
        return ans;
    }

    private void dfs(TreeNode root, int l, int r) {
        if (root == null) {
            return;
        }
        ans = Math.max(ans, Math.max(l, r));
        dfs(root.left, r + 1, 0);
        dfs(root.right, 0, l + 1);
    }
}


// Solution 2
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private static class Frame {
        TreeNode node;
        int l;
        int r;

        Frame(TreeNode node, int l, int r) {
            this.node = node;
            this.l = l;
            this.r = r;
        }
    }

    public int longestZigZag(TreeNode root) {
        int ans = 0;
        Deque<Frame> stk = new ArrayDeque<>();
        if (root != null) {
            stk.push(new Frame(root, 0, 0));
        }
        while (!stk.isEmpty()) {
            Frame cur = stk.pop();
            ans = Math.max(ans, Math.max(cur.l, cur.r));
            if (cur.node.right != null) {
                stk.push(new Frame(cur.node.right, 0, cur.l + 1));
            }
            if (cur.node.left != null) {
                stk.push(new Frame(cur.node.left, cur.r + 1, 0));
            }
        }
        return ans;
    }
}


  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int ans = 0;

    int longestZigZag(TreeNode* root) {
        dfs(root, 0, 0);
        return ans;
    }

    void dfs(TreeNode* root, int l, int r) {
        if (!root) return;
        ans = max(ans, max(l, r));
        dfs(root->left, r + 1, 0);
        dfs(root->right, 0, l + 1);
    }
};


// Solution 2
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int longestZigZag(TreeNode* root) {
        int ans = 0;
        vector<tuple<TreeNode*, int, int>> stk{{root, 0, 0}};
        while (!stk.empty()) {
            auto [node, l, r] = stk.back();
            stk.pop_back();
            if (!node) {
                continue;
            }
            ans = max(ans, max(l, r));
            stk.emplace_back(node->right, 0, l + 1);
            stk.emplace_back(node->left, r + 1, 0);
        }
        return ans;
    }
};


  • # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def longestZigZag(self, root: TreeNode) -> int:
        def dfs(root, l, r):
            if root is None:
                return
            nonlocal ans
            ans = max(ans, l, r)
            dfs(root.left, r + 1, 0)
            dfs(root.right, 0, l + 1)

        ans = 0
        dfs(root, 0, 0)
        return ans



# Solution 2
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def longestZigZag(self, root: TreeNode) -> int:
        ans = 0
        stk = [(root, 0, 0)]
        while stk:
            node, l, r = stk.pop()
            if node is None:
                continue
            ans = max(ans, l, r)
            stk.append((node.right, 0, l + 1))
            stk.append((node.left, r + 1, 0))
        return ans


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func longestZigZag(root *TreeNode) int {
	ans := 0
	var dfs func(root *TreeNode, l, r int)
	dfs = func(root *TreeNode, l, r int) {
		if root == nil {
			return
		}
		ans = max(ans, max(l, r))
		dfs(root.Left, r+1, 0)
		dfs(root.Right, 0, l+1)
	}
	dfs(root, 0, 0)
	return ans
}


// Solution 2
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func longestZigZag(root *TreeNode) int {
	ans := 0
	type frame struct {
		node *TreeNode
		l, r int
	}
	stk := []frame{{root, 0, 0}}
	for len(stk) > 0 {
		cur := stk[len(stk)-1]
		stk = stk[:len(stk)-1]
		if cur.node == nil {
			continue
		}
		ans = max(ans, max(cur.l, cur.r))
		stk = append(stk, frame{cur.node.Right, 0, cur.l + 1}, frame{cur.node.Left, cur.r + 1, 0})
	}
	return ans
}

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