236. Lowest Common Ancestor of a Binary Tree

[Tcdian]

236. Lowest Common Ancestor of a Binary Tree

给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉树:  root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note

• 所有节点的值都是唯一的。
• p、q 为不同节点且均存在于给定的二叉树中。

[Tcdian]

Solution

• JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
    let pPath;
    let qPath;
    const path = [];
    dfs(root);
    let result;
    for (let i = 0; i < pPath.length; i++) {
        if (pPath[i] !== qPath[i]) {
            break;
        }
        result = pPath[i];
    }
    return result;
    
    function dfs(root) {
        if (pPath && qPath) {
            return;
        }
        path.push(root);
        if (root === p) {
            pPath = [...path];
        }
        if (root === q) {
            qPath = [...path];
        }
        if (root.left) {
            dfs(root.left);
            path.pop();
        }
        if (root.right) {
            dfs(root.right);
            path.pop();
        }
    }
};