[LeetCode][JavaScript]Lowest Common Ancestor of a Binary Tree

Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

 

 

 

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比较难以描述...

我的理解是把目标的节点向上"冒泡"，拿这个做例子：the lowest common ancestor (LCA) of nodes 5 and 1 is 3

从下往上看，6这个节点往上一层返回null，因为这个节点本身，它的左子树和右子树都不包含1和5。

同理7，4，2返回null，

5这个节点左右子树不包含1和5，但它自身是5，返回5，

同理以1为根节点的树返回1，

以1为根节点的树发现左右都不为null，说明找到了。

以上是一种情况，还有一种情况就是一旦发现目前的节点等于p或者q，就可以直接返回，因为默认输入是正确的。

/**
 * 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) {
    if(root === null || root === p || root === q){
        return root;
    }
    var left = lowestCommonAncestor(root.left, p, q);
    var right = lowestCommonAncestor(root.right, p, q);
    return left !== null && right !== null ? root : (left !== null ? left : right);
}