积少成多

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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.

===============

CLA问题,

利用递归方式:采用的后序遍历的思想,LRN

左子树查找,右子树查找,这个思路进行.

TreeNode *seek(TreeNode *root,TreeNode *p,TreeNode *q){
 help_seek();  
}
void help_seek(TreeNode *root, TreeNode *p,TreeNode *q,bool l,bool r){
 if(r && l) cout<<"find the value""<<root->val<<endl;
 if(root){
   help_seek(root->left);
   help_seek(root->right); 
 }
}

 

???????

====

代码:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==nullptr) return NULL;
        if(root==p || root==q) return root;
        TreeNode *left_lca = lowestCommonAncestor(root->left,p,q);
        TreeNode *right_lca = lowestCommonAncestor(root->right,p,q);
        if(left_lca&&right_lca) return root;

        return left_lca==nullptr? right_lca:left_lca;
    }
};

 

posted on 2016-06-26 21:01  x7b5g  阅读(169)  评论(0编辑  收藏  举报